OK so I'm trying to understand why the angle of a pendulum as a function of time is a sine wave.

I can't really find an explanation online and when I do find something partial there are certain symbols I don't understand.

$$\frac{d^2\theta}{dt^2} + \frac{g}{l}\sin\theta = 0$$

This is the equation I found on Wikipedia.

  1. What I don't understand here is the part $d^2\theta\over dt^2$, from what I know $d$ means instantaneous delta, instantaneous rate of change, why does the upper part of the function has the square sign right after the $d$ ($d^2\theta$) and the lower part of the fraction has the square sign after the $t$ and not after the $d$ ($dt^2$).

  2. This part still doesn't really show the answer to my question cause the change of the angle and the time are squared so it doesn't mean much to, I'm hoping for an explanation that is simple as possible and intuitive as possible for why the angle as a function of time is a sine wave, hopefully as much mechanics as possible ans less math, a good reference is also good.

EDIT: OK, I Understand the first part (the square sign notation), what I still don't understand is how we can get from the the equation I wrote above, or from the acceleration as function of the angle: $a = g\sin\ \theta$, to an equation that shows the angle as a function of time.

It's confusing for me since the angle itself changes all the time, in both eqaution (the one at the top and $a = g\sin\ \theta$ we have $\theta$, but $\theta$ itself changes all the time!

It seems it's not possible to use "little math" here, so use math where necessary.

Thank you.

  • 2
    $\begingroup$ Hi fiftyeight, and welcome to Physics Stack Exchange! It sounds like the root of your problem is that you haven't learned calculus, in particular that you haven't learned about derivatives. In that case I wouldn't worry too much about the mathematical aspects of this (like the meaning of $\frac{\mathrm{d}^2}{\mathrm{d}t^2}$); you'll get there eventually but this site is not the place to learn calculus from scratch. Someone should still be able to give you an intuitive explanation of why a pendulum's angle approximately follows a sine wave, though. $\endgroup$ – David Z Dec 19 '11 at 4:03
  • $\begingroup$ I know what derivatives are and the basics of calculus but just never seen what I described in question 1. I understand what a second derivative, I just find it strange that the square sign is before the $\theta$ but after the $t$ Thank you for the warm welcome :) $\endgroup$ – fiftyeight Dec 19 '11 at 4:30
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    $\begingroup$ It is also worth to note that with the equation given you will not obtain sine wave. $\endgroup$ – Misha Dec 19 '11 at 4:55
  • $\begingroup$ Really, you've never seen a second derivative written in this notation? Sounds like your teachers have done you a disservice ;-) Anyway, the reason it's $\mathrm{d}^2\theta$ instead of $\mathrm{d}\theta^2$ is a mathematical thing. That's what I'm telling you not to worry about; it will make more sense when you understand derivative operators better. $\endgroup$ – David Z Dec 19 '11 at 5:31

$\frac{\mathrm{d}^2\theta}{\mathrm{d}t^2}$ is the second time derivative of angular displacement. $\frac{\mathrm{d}\theta}{\mathrm{d}t}$ would be first time derivative. In order to understand this displacement, let's compare it with linear displacement $x$

$\frac{\mathrm{d}x}{\mathrm{d}t}$ is speed, while $\frac{\mathrm{d}^2x}{\mathrm{d}t^2}$ is acceleration. So if the first time derivative is the rate of change of a quantity with respect to time, the second derivative measures the rate of change of that rate of change!

With this in mind, if you look at the "Force derivation" on the same page it shows how you can use acceleration (second derivative with respect to time) to derive the pendulum differential equation. It also shows the origin of $\sin\theta$ dependence, which comes from resolving the gravitational force into two perpendicular components. The $\sin\theta$ component is tangential to the arc traced out by the motion of the pendulum and the only one relevant to calculating the change in speed.

Also, to answer your question regarding the placement of "2" in the notation, you should think of $\mathrm{d}()/\mathrm{d}t$ as an operator that acts upon any function placed inside the $()$. So $\mathrm{d}^2/\mathrm{d}t^2$ can be written out as $\frac{\mathrm{d}}{\mathrm{d}t}(\frac{\mathrm{d}}{\mathrm{d}t}())$.

Edit to show the solution as requested by the OP: If you would like to actually see the solution then here is one approximation. To make our lives easier, we need take this second-order non-linear differential equation and make it linear. This is achieved using the small angle approximations $\sin(\theta)\sim\theta$ mentioned by @MarkEichenlaub.

We have:

$\frac{\mathrm{d^2}\theta}{\mathrm{d}t^2} + \frac{g}{l}\theta = 0$

The solution to such an equation will be proportional to $e^{\lambda t}$, where $\lambda$ is a constant. Substitute that into the equation:

$\frac{\mathrm{d^2}(e^{\lambda t})}{\mathrm{d}t^2} + \frac{g}{l}(e^{\lambda t}) = 0$

For brevity I am leaving out a few steps, but if you work through it you should end up with two solutions, the sum of which will be the general solution.

$\theta_1 = c_1 \times exp(-it\sqrt{g/l})$ and $\theta_2 = c_2 \times exp(it\sqrt{g/l})$, where $i$ is an imaginary number and it appears because we take a square root of a negative number. Once again, after omitting a few steps and using Euler's identity, we end up with the general solution (the sum of the two solutions) as

$\theta = c_1\times\cos(t\sqrt{g/l}) + c_2\times\sin(t\sqrt{g/l})$ and there you have $\theta$ on one side and $t$ on the other. I'm afraid you will have spend some time working through the maths to get where and how we arrive at this solution. Also, it is valid as long as the small angle approximation is valid.

  • $\begingroup$ Okay, I can see more or less why the acceleration in the direction tangent to the arc depends on $sin\ \theta$ but how does this lead to the fact that the angular displacement itself is dependant on $sin\ \theta$, and is angular displacement another name for $\Delta \theta$ $\endgroup$ – fiftyeight Dec 19 '11 at 4:38
  • $\begingroup$ @fiftyeight $\sin\theta$ results from decomposing gravitational force into two components. On the above referenced page, if you look at Figure 1, you can see a right-angled triangle formed by $mg$ (in blue), and purple vector ($mg\cos\theta$). Using trigonometric relations, the tangential component is $mg\sin\theta$, hence the appearance of $\sin\theta$. $\endgroup$ – Omar Dec 19 '11 at 14:46
  • $\begingroup$ OK, I can see why the acceleration in the direction perpendicular to the arc is $g\sin\ \theta$, what I find most confusing is that $\theta$ itself changes all the time so I don't see how $g\sin\ \theta$ really helps me. It seems that it's a must to use math here so feel welcome to use math. i'll edit my qeustion to reflect this. $\endgroup$ – fiftyeight Dec 19 '11 at 16:44
  • $\begingroup$ @fiftyeight I have edited the answer to show one solution to the differential equation. I am not convinced it will make things any more intuitive for you, but might be good for you to work through the solution. $\endgroup$ – Omar Dec 19 '11 at 20:12
  • $\begingroup$ I can tell you without doubt that you were right about one thing: "you will have spend some time working through the maths to get where and how we arrive at this solution", I'll work through it, can someone just tell me what topic I need to look up to understand why "The solution to such an equation will be proportional to $e^{\lambda t}$" $\endgroup$ – fiftyeight Dec 19 '11 at 22:15

The answer is that you'll have to learn more math to understand it. Here is a rough explanation:

Think of a string hanging from the ceiling with a weight at the bottom. That's a pendulum, but it doesn't have to swing in a plane the way a grandfather clock's pendulum does. It can move freely, and so it can also go around in little circles.

Imagine two of these things next to each other. The one on the left is going in small circles with a 1 cm radius. The one on the right is swinging back and forth in a plane with a 1 cm amplitude. Aside from that, though, they keep perfect time with each other.

The left/right part of a circle is a cosine function; that's the definition of cosine. So, since the two pendulums keep the same time, the planar one shakes back and forth as a cosine function, too. (Your question asked for why it's a sine function, but these are essentially the same, and equally-valid solutions.)

Note that this only works for small angles. At larger angles, the circular pendulum and the left/right pendulum don't keep perfect time with each other.

The essential reason this asynchrony occurs is that at larger angles the tension in the string changes significantly throughout a cycle as the pendulum swings back and forth, but in a pendulum going in a circle the tension is constant.

So in fact, a sine function is not the precise solution to the mathematical equation you provided. It is only an approximate solution.

  • $\begingroup$ +1 Great answer. You brought in the small-angle approximation. From there you showed 1) for the linear pendulum the result is not sine(or cosine) as it goes beyond a small angle, and 2) for the circular pendulum it's even less so (because the pendulum could be set swinging around in a circle at high speed). $\endgroup$ – Mike Dunlavey Dec 19 '11 at 17:56
  • $\begingroup$ Thank you, but I edited my question cause it seems everyone was going a little too easy on me in the math area. $\endgroup$ – fiftyeight Dec 19 '11 at 18:51

That equation is equal to

$$\frac{d^2\theta}{dt^2} = -\frac{g}{l}\sin\theta$$

which is the same as:

$$\frac{d}{dt}(\frac{d\theta}{dt})= -\frac{g}{l}\sin\theta$$

(That's where the notation comes from.) This is only sinusoidal to the extent that $\theta$ is near zero, where $\sin\theta = \theta$.

If the equation were:

$$\frac{d}{dt}(\frac{d\theta}{dt})= -\frac{g}{l}\theta$$

then $\theta(t)$ would be exactly sinusoidal, as you could prove by substitution.

EDIT: OK, unless you can integrate directly, you can guess a solution to the ODE and try it. Suppose $\theta(t) = a\sin(bt)$. Then the first derivative wrt $t$ is $ab\cos(bt)$ The second derivative is $-ab^2\sin(bt)$. So you replace and you get

$$-ab^2\sin(bt) = -\frac{g}{l}a\sin(bt)$$

which simplfies to

$$b^2 = \frac{g}{l}$$

and solving for $b$ makes both side equal. So by assuming $\theta$ was sinusoidal, we made the differential equation balance.

  • $\begingroup$ I still don't see it, how do you get from one of those equations to an equation where you have $\theta$ on one side, and $t$ on the other? $\endgroup$ – fiftyeight Dec 19 '11 at 19:00
  • $\begingroup$ @fiftyeight: Does that edit help? It could be solved by direct integration, but I was taught the "guessing" method, which works when it's not easy to integrate. $\endgroup$ – Mike Dunlavey Dec 19 '11 at 21:16

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